In ΔABC, points D and E are on sides AB and AC respectively, with DE parallel to BC. Given that: - AD = 3 cm - DB = 2 cm - DE || BC According to the Basic Proportionality Theorem (BPT): - When a line is drawn parallel to one side of a triangle intersecting the other two sides, it divides those sidesRead more
In ΔABC, points D and E are on sides AB and AC respectively, with DE parallel to BC.
Given that:
– AD = 3 cm
– DB = 2 cm
– DE || BC
According to the Basic Proportionality Theorem (BPT):
– When a line is drawn parallel to one side of a triangle intersecting the other two sides, it divides those sides in the same ratio
– Therefore, AD:DB = AE:EC = 3:2
While we know the ratio AE:EC = 3:2, we cannot determine the actual length of AE because:
1. The total length of AC is unknown
2. Without knowing AC, we cannot split it in the ratio 3:2 to find AE
3. Having just the ratio 3:2 and no information about the total length AC means there could be infinitely many possible values for AE
For example:
– If AC = 10 cm, then AE would be 6 cm
– If AC = 15 cm, then AE would be 9 cm
– If AC = 5 cm, then AE would be 3 cm
Therefore, the length of AE cannot be determined with the given information.
When two triangles are similar (ΔABC ~ ΔDEF), their areas and sides follow a particular mathematical relationship: If area ratio = m:n, then side ratio = √m:√n Given: - ΔABC ~ ΔDEF - ar(ABC):ar(DEF) = 16:25 Therefore: 1. The side ratio is obtained by square root of the area ratio 2. Side ratio = √16Read more
When two triangles are similar (ΔABC ~ ΔDEF), their areas and sides follow a particular mathematical relationship:
If area ratio = m:n, then side ratio = √m:√n
Given:
– ΔABC ~ ΔDEF
– ar(ABC):ar(DEF) = 16:25
Therefore:
1. The side ratio is obtained by square root of the area ratio
2. Side ratio = √16:√25
3. Simplifying: 4:5
AB:DE = 4:5
This relationship holds because:
– Area ratio = (Side ratio)²
– Suppose side ratio = x:y, then area ratio = x²:y²
– In a similar vein, if area ratio = m:n, then side ratio = √m:√n
– Now, in this example, √16:√25 = 4:5
The above mathematical equivalence applies to every pair of similar triangles because area ratio is always equal to the square of ratio of the respective sides.
In two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides. Given that the ratio of the corresponding sides is 4:9, the ratio of their areas can be calculated as: (Ratio of areas ) = ( Ratio of sides )² = ( 4:9 )² = 4² : 9² = 16:81 Thus, the cRead more
In two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Given that the ratio of the corresponding sides is 4:9, the ratio of their areas can be calculated as:
(Ratio of areas ) = ( Ratio of sides )²
= ( 4:9 )²
= 4² : 9²
= 16:81
If in ΔABC, D and E are points on sides AB and AC respectively such that DE || BC and AD = 3 cm, DB = 2 cm, then AE equals:
In ΔABC, points D and E are on sides AB and AC respectively, with DE parallel to BC. Given that: - AD = 3 cm - DB = 2 cm - DE || BC According to the Basic Proportionality Theorem (BPT): - When a line is drawn parallel to one side of a triangle intersecting the other two sides, it divides those sidesRead more
In ΔABC, points D and E are on sides AB and AC respectively, with DE parallel to BC.
Given that:
– AD = 3 cm
– DB = 2 cm
– DE || BC
According to the Basic Proportionality Theorem (BPT):
– When a line is drawn parallel to one side of a triangle intersecting the other two sides, it divides those sides in the same ratio
– Therefore, AD:DB = AE:EC = 3:2
While we know the ratio AE:EC = 3:2, we cannot determine the actual length of AE because:
1. The total length of AC is unknown
2. Without knowing AC, we cannot split it in the ratio 3:2 to find AE
3. Having just the ratio 3:2 and no information about the total length AC means there could be infinitely many possible values for AE
For example:
– If AC = 10 cm, then AE would be 6 cm
– If AC = 15 cm, then AE would be 9 cm
– If AC = 5 cm, then AE would be 3 cm
Therefore, the length of AE cannot be determined with the given information.
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If ΔABC ~ ΔDEF, and ar(ABC):ar(DEF) = 16:25, then AB:DE equals:
When two triangles are similar (ΔABC ~ ΔDEF), their areas and sides follow a particular mathematical relationship: If area ratio = m:n, then side ratio = √m:√n Given: - ΔABC ~ ΔDEF - ar(ABC):ar(DEF) = 16:25 Therefore: 1. The side ratio is obtained by square root of the area ratio 2. Side ratio = √16Read more
When two triangles are similar (ΔABC ~ ΔDEF), their areas and sides follow a particular mathematical relationship:
If area ratio = m:n, then side ratio = √m:√n
Given:
– ΔABC ~ ΔDEF
– ar(ABC):ar(DEF) = 16:25
Therefore:
1. The side ratio is obtained by square root of the area ratio
2. Side ratio = √16:√25
3. Simplifying: 4:5
AB:DE = 4:5
This relationship holds because:
– Area ratio = (Side ratio)²
– Suppose side ratio = x:y, then area ratio = x²:y²
– In a similar vein, if area ratio = m:n, then side ratio = √m:√n
– Now, in this example, √16:√25 = 4:5
The above mathematical equivalence applies to every pair of similar triangles because area ratio is always equal to the square of ratio of the respective sides.
Click here for more:
See lesshttps://www.tiwariacademy.in/ncert-solutions-class-10-maths-chapter-6/
If in two similar triangles, the ratio of their corresponding sides is 4:9, then the ratio of their areas is:
In two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides. Given that the ratio of the corresponding sides is 4:9, the ratio of their areas can be calculated as: (Ratio of areas ) = ( Ratio of sides )² = ( 4:9 )² = 4² : 9² = 16:81 Thus, the cRead more
In two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Given that the ratio of the corresponding sides is 4:9, the ratio of their areas can be calculated as:
(Ratio of areas ) = ( Ratio of sides )²
= ( 4:9 )²
= 4² : 9²
= 16:81
Thus, the correct answer is 16:81.
Click here for more:
See lesshttps://www.tiwariacademy.in/ncert-solutions-class-10-maths-chapter-6/